On 16-08-2019 09:01, Bruce Kellett wrote:
On Fri, Aug 16, 2019 at 4:43 PM smitra <smi...@zonnet.nl> wrote:
On 16-08-2019 06:31, Bruce Kellett wrote:
On Wed, Aug 14, 2019 at 8:28 AM smitra <smi...@zonnet.nl> wrote:
On 13-08-2019 13:33, Bruce Kellett wrote:
Of course A(x) and B(x) refer to the same point on the screen.
That is
not a collapse, that is just what the notation means.
A(x) and B(x) considered as the representations of |A> and |B>
in
the
position basis, i.e. A(x) = <x|A> and B(x) = <x|B> are still
orthogonal
states, as they represent the orthogonal states |A> and |B>:
0 = <A|B> = Integral over x of <A|x><x|B>d^3x = Integral over x
of
A*(x)B(x) d^3x
I don't think this really works out. You are claiming that the
integral of the interference terms over the whole screen
vanishes. If
we look at the usual derivation of the interference from two
slits, we
get something like
Intensity I = 2 A^2 (sin^2(beta)/beta^2) (1 + cos(delta))
where the term involving the angle beta is the superposed
diffraction
pattern from the finite width of the slits. The cos (delta) term
is
the interference, but it has this form only in a small angle
approximation, and the phase difference delta is, of course,
limited
by the separation of the slits. So, although the cos(delta) term
may
integrate to zero over small angles, the presence of the
diffraction
envelope, and the limitations of the small angle approximation,
mean
that is almost certainly will not vanish when integrated over the
whole screen.
Yes, this is in the small angle approximation, if you go beyond
that
then the itnegral over the screen won't vanish.
So <A|B> will not vanish in general.
No, because <A|B> is the integral over all space and this is
exactly
zero. What happens is that when the small angle approximation
becomes
invalid and the integral over only the screen becomes nonzero, the
integrals over surfaces parallel to the screen will have a values
that
differ by a phase factor that depends on the distance in the
direction
orthogonal to the screen. This then causes the integral over all
space
to vanish.
I think you need to prove that. In my understanding, A(x) = <x|A> is
to be interpreted as the amplitude for a wave through slit A to get to
the screen at x. There is nothing 3-dimensional about this. The 'x' is
just the distance from the centre of the screen in the plane of the
screen. Nothing else is relevant. You do not have to integrate over
all space because you use a complete set of states in the x-direction:
int |x><x| dx = 1.
A quantum state is defined in the position representation by assigning
an amplitude to all points in space. It can be the case that the
amplitude is zero outside of a narrow volume surrounding the the screen,
in which case the integration can be approximated as an integral over
the screen's surface.
The orthogonality can be rigorously proved as follows. If we have a
single particle incident on the two slits described by a time dependent
wave function psi(x,t) = 1/sqrt(2) [A(x,t) + B(x,t)] such that at A(x,0)
is nonzero at one slit and B(x,0) at the other slit, then A(x,0) and
B(x,0) are obviously orthogonal. Since time evolution will preserve
inner products, A(x,t) and B(x,t) will remain orthogonal as a function
of t. One can then describe the interaction with the screen as an
effective collapse that will happen with the largest probability when
the peak of the wavefunction has arrived at the screen.
Saibal
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